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Construct a triangle ABC with side BC = 7 cm, $\angle B=45^{\circ},\angle A=105^{\circ}$ . Then construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of the $\bigtriangleup ABC$ .

Answers (1)

Given : In $\bigtriangleup ABC$

$BC=7\; cm$

$\angle B=45^{\circ}$

$\angle A=105^{\circ}$

$\angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow 105^{\circ}+45^{\circ}+\angle C=180^{\circ}$

$\Rightarrow \angle C=30^{\circ}$

Steps for construction :

(i) Draw a line BC of 7 cm.

(ii) Draw $\angle B=45^{\circ}$ and $\angle C=30^{\circ}$

(iii) Let point A be the point where rays from B and C intersect.

(iv) $\bigtriangleup ABC$ is the required triangle.

(v) Now take scale factor $\frac{3}{4}<1$ .

(vi) Make a ray BX making acute angle with BC.

(vii) Mark 4 points $(B_1,B_2,B_3,B_4)$ on BX such that $BB_1=B_1B_2=B_2B_3=B_3B_4$ .

(viii) Join $B_4C$ .

(ix) Draw a line parallel to $B_4C$ from point $B_3$ intersecting BC at $C'$ . Join $B_3C'$ .

(x) Draw a parallel line $(A'C')$ to AC intersecting at $A'$

(xi) $\bigtriangleup A'BC'$ is formed which $\frac{3}{4}\bigtriangleup ABC$ .

Posted by

Ravindra Pindel

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